Finding Subsets of a Cartesian Product

Let's break down this problem about finding the number of subsets of a Cartesian product with a certain number of elements. We are given two sets, A and B, with a specific number of elements.

• Set A contains 2 elements. We can write this as $|A| = 2$.
• Set B contains 4 elements. We can write this as $|B| = 4$.

We need to find the number of subsets of the Cartesian product $A \times B$ that have 3 or more elements.

Understanding the Cartesian Product \(A \times B\)

The Cartesian product $A \times B$ is the set of all possible ordered pairs where the first element comes from set A and the second element comes from set B. The number of elements in the Cartesian product is the product of the number of elements in each set.

So, the number of elements in $A \times B$ is:

\(|A \times B| = |A| \times |B|\)

\(|A \times B| = 2 \times 4 = 8\)

The set $A \times B$ has 8 elements.

Calculating the Total Number of Subsets

A set with \(n\) elements has a total of \(2^n\) subsets. The set $A \times B$ has 8 elements, so the total number of subsets of $A \times B$ is \(2^8\).

Total number of subsets = \(2^8 = 256\)

Finding Subsets with 3 or More Elements

We are interested in subsets that have 3 or more elements. This means subsets with exactly 3 elements, exactly 4 elements, ..., up to exactly 8 elements. Calculating each of these individually using combinations and then summing them up would be lengthy.

A simpler approach is to find the total number of subsets and subtract the number of subsets that do not meet the condition (i.e., those with less than 3 elements).

Subsets with less than 3 elements are those with:

• 0 elements (the empty set)
• 1 element
• 2 elements

The number of subsets of a set with \(n\) elements having exactly \(k\) elements is given by the binomial coefficient \(\binom{n}{k}\), which is calculated as \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\).

In our case, \(n = |A \times B| = 8\). We need to calculate the number of subsets with \(k=0\), \(k=1\), and \(k=2\) elements.

Number of Subsets with 0 Elements

This is the number of subsets with exactly 0 elements from a set of 8 elements:

\(\binom{8}{0} = \frac{8!}{0!(8-0)!} = \frac{8!}{0!8!} = \frac{8!}{1 \times 8!} = 1\)

There is only 1 subset with 0 elements (the empty set).

Number of Subsets with 1 Element

This is the number of subsets with exactly 1 element from a set of 8 elements:

\(\binom{8}{1} = \frac{8!}{1!(8-1)!} = \frac{8!}{1!7!} = \frac{8 \times 7!}{1 \times 7!} = 8\)

There are 8 subsets with 1 element.

Number of Subsets with 2 Elements

This is the number of subsets with exactly 2 elements from a set of 8 elements:

\(\binom{8}{2} = \frac{8!}{2!(8-2)!} = \frac{8!}{2!6!} = \frac{8 \times 7 \times 6!}{ (2 \times 1) \times 6!} = \frac{8 \times 7}{2} = 4 \times 7 = 28\)

There are 28 subsets with 2 elements.

Total Number of Subsets with Less Than 3 Elements

The total number of subsets with 0, 1, or 2 elements is the sum of the numbers calculated above:

Number of subsets (< 3 elements) = \(\binom{8}{0} + \binom{8}{1} + \binom{8}{2}\)

Number of subsets (< 3 elements) = \(1 + 8 + 28 = 37\)

Number of Subsets with 3 or More Elements

Finally, to find the number of subsets with 3 or more elements, we subtract the number of subsets with less than 3 elements from the total number of subsets.

Number of subsets (\(\ge\) 3 elements) = Total number of subsets - Number of subsets (< 3 elements)

Number of subsets (\(\ge\) 3 elements) = \(256 - 37\)

Number of subsets (\(\ge\) 3 elements) = \(219\)

Thus, there are 219 subsets of \(A \times B\) having 3 or more elements.

Revision Table: Key Concepts

Additional Information: Power Sets and Combinations

The collection of all subsets of a set S is called the power set of S, denoted by P(S). The number of elements in the power set is \(|P(S)| = 2^{|S|}\).

Combinations, denoted by \(\binom{n}{k}\) or \(C(n, k)\), are used when we want to choose \(k\) elements from a set of \(n\) elements without regard to the order of selection. This is exactly what we do when forming subsets of a specific size \(k\).

The total number of subsets \(2^n\) can also be seen as the sum of the number of subsets of each possible size:

\(2^n = \binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \dots + \binom{n}{n}\)

This is the binomial theorem applied to \((1+1)^n\). In this problem, we used this principle in reverse: we calculated the total and subtracted the terms we didn't want (\(\binom{8}{0}, \binom{8}{1}, \binom{8}{2}\)) to find the sum of the remaining terms (\(\binom{8}{3} + \dots + \binom{8}{8}\)).

This method of finding the complement (subtracting the unwanted cases from the total) is a common and efficient technique in combinatorics and set theory problems.